Question 1147657
let T = the tens digit.
let U = the units digit.
the number will be 10T + U.
the new number with the digits reversed will be 10U + T.
you are given that the units digits is one more than the tens digit.
this means that U = T + 1
you are given that the new number is 3 less than twice the original number.
this means that 10U + T = 2 * (10T + U) - 3
your two equations that need to be solved simultaneously are:
U = T + 1
10U + T = 2 * (10T + U) - 3
simplify the second equation and leave the first equation as is to get:
U = T + 1
10U + T = 20T + 2U - 3
place the variables on the left side of each equation and the constants on the right side of each equation to get:
U - T = 1
8U - 19T = -3
multiply both sides of the first equation by -8 and leave the second equation as is to get:
-8U + 8T = -8
8U - 19T = -3
add both equations together to get:
-11T = -11
solve for T to get:
T = 1
since U = T + 1, then U = 2
the number is 12.
the new number with the digits reversed is 21.
the units digits is 1 more than the tens digits.
this is true, since 2 is 1 more than 1.
the reversed number is 3 less than two times the original number.
this is true, since 21 = 2 * 12 - 3 = 24 - 3 = 21
solution looks good.
solution is that the original number is 12.